I wish to prove that if we have a linear differential equation of order n with constant coefficients:
$$\frac{d^n}{dt^n}z(t)+a_{n-1}\frac{d^{n-1}}{dt^{n-1}}z(t)+\cdots+a_1\frac{d}{dt}z(t)+a_0z(t)=0 \quad(1) $$
If $\lambda$ is a root of the characteristic polynomial with multiplicity m, so a set of fundemental solutions will be:
$\phi_k:\mathbb{R} \rightarrow \mathbb{R}, t\rightarrow t^k e^{\lambda t},\text{with}\quad k=\{0,\cdots,m-1\}$.
Defining:
$ \begin{aligned} D^{k}: \quad \mathcal{C}^{\infty}(\mathbb{R}, \mathbb{K}) & \longrightarrow \quad \mathcal{C}^{\infty}(\mathbb{R}, \mathbb{K}) \\ \phi & \mapsto \frac{d^{k} \phi}{d t^{k}} \end{aligned}$
Defining P(D) has the characteristic polynomial: P(D)z=0. I can suppose that:
$P(D)\left(t^{q} e^{\lambda t}\right)=\sum_{k=0}^{n} a_{k} \sum_{j=0}^{k}\left(\begin{array}{c} k \\ j \end{array}\right) D^{j}\left(t^{q}\right) D^{k-j}\left(e^{\lambda t}\right)=\sum_{k=0}^{n} a_{k}\left(\sum_{j=0}^{q}\left(\begin{array}{c} k \\ j \end{array}\right) D^{j}\left(t^{q}\right) \lambda^{k-j} e^{\lambda t}\right)$
And do not have a clue how to finish the proof.